Answer :
To solve the given expression [tex]\(\frac{7^{13}}{7}\)[/tex], we can use the laws of exponents. Specifically, we will use the rule that states:
[tex]\[ \frac{a^m}{a^n} = a^{m-n} \][/tex]
Here, [tex]\(a\)[/tex] is the base (which is 7 in our case), [tex]\(m\)[/tex] is the exponent in the numerator (which is 13), and [tex]\(n\)[/tex] is the exponent in the denominator (which is 1, since [tex]\(7\)[/tex] can be written as [tex]\(7^1\)[/tex]).
Applying the rule:
[tex]\[ \frac{7^{13}}{7} = \frac{7^{13}}{7^1} = 7^{13-1} = 7^{12} \][/tex]
So, the expression [tex]\(\frac{7^{13}}{7}\)[/tex] simplifies to [tex]\(7^{12}\)[/tex].
The equivalent expression to [tex]\(7^{12}\)[/tex] is not directly listed in the given options (A: [tex]\(7^{20}\)[/tex], B: [tex]\(7^5\)[/tex], C: [tex]\(7^{91}\)[/tex], D: [tex]\(7^6\)[/tex]). Therefore, it appears that the correct simplified form of the expression does not match any of the provided choices.
However, the simplified expression is:
[tex]\[ 7^{12} \][/tex]
The equivalent numerical value here, coming from the simplified exponent, is [tex]\(12\)[/tex].
[tex]\[ \frac{a^m}{a^n} = a^{m-n} \][/tex]
Here, [tex]\(a\)[/tex] is the base (which is 7 in our case), [tex]\(m\)[/tex] is the exponent in the numerator (which is 13), and [tex]\(n\)[/tex] is the exponent in the denominator (which is 1, since [tex]\(7\)[/tex] can be written as [tex]\(7^1\)[/tex]).
Applying the rule:
[tex]\[ \frac{7^{13}}{7} = \frac{7^{13}}{7^1} = 7^{13-1} = 7^{12} \][/tex]
So, the expression [tex]\(\frac{7^{13}}{7}\)[/tex] simplifies to [tex]\(7^{12}\)[/tex].
The equivalent expression to [tex]\(7^{12}\)[/tex] is not directly listed in the given options (A: [tex]\(7^{20}\)[/tex], B: [tex]\(7^5\)[/tex], C: [tex]\(7^{91}\)[/tex], D: [tex]\(7^6\)[/tex]). Therefore, it appears that the correct simplified form of the expression does not match any of the provided choices.
However, the simplified expression is:
[tex]\[ 7^{12} \][/tex]
The equivalent numerical value here, coming from the simplified exponent, is [tex]\(12\)[/tex].
